T = rF where F = mg
Tnet = T1 – T2 = 0
T1 = T2
rF = rF =>
Unknown mass = (M2*x2)/x1
Mass:
D = .29kg % Error – 7.4 %
F = .22kg % Error – 15.8%
R = .15kg % Error - 15.4%
B = .28kg % Error – 7.7%
E = .22kg % Error – 4.8%
Case IV:
m1*x1 = m2*x2 where m1 = mass of meter stick
Assume weight acts at center of mass of meter stick. Assume 50cm mark is center of mass.
m1 = (m2*x2)/x1 = (.1kg*.26m)/.24 = .11kg
Actual mass of meter stick - .11kig
% Error = 0%
Questions:
A large or small force can produce a large or small torque depending on the value of r-perpendicular. If a small force is applied at a 90 degree angle, it can easily produce a higher torque than a large force applied at at smaller angle.
We do not include the “support force” in our calculations because its r value, or the distance from the axis is zero, causing it to produce no Torque.
Case III:
T = rF where F = mg
Tnet = T1 – T2 = 0
T1 = T2
rF = rF =>
Unknown mass = (M2*x2)/x1
Mass:
D = .29kg % Error – 7.4 %
F = .22kg % Error – 15.8%
R = .15kg % Error - 15.4%
B = .28kg % Error – 7.7%
E = .22kg % Error – 4.8%
Case IV:
m1*x1 = m2*x2 where m1 = mass of meter stick
Assume weight acts at center of mass of meter stick. Assume 50cm mark is center of mass.
m1 = (m2*x2)/x1 = (.1kg*.26m)/.24 = .11kg
Actual mass of meter stick - .11kig
% Error = 0%
Questions:
(m1*r1 – m2*r2)/rs = ms = .17kg